Question

the region between x=0, x=8 and bounded by the graphs of y=x^1/3 and y=10e^-0.1x is rotated around the x axis to form a solid. Find its exact volume

Answer 1

so bounds 0 to 8 pi r ^2

Answer 2

First graph both functions in order to determine which one is the function on the top bound and which one is the function that forms the bottom bound. And then use the washer method to solve for the area. Using the form\[\pi \int\limits _0 ^8 f(x)^2 - g(x)^2 dx\] where f(x) is the "top" function and g(x) is the "bottom" function.

Answer 3

yeah i got that then what

Answer 4

Just simplify the integral and you have your answer :p

Answer 5

do i not include the 10e^-.1x?

Answer 6

Yes...you find out if it is the top or bottom bound and then you plug it in for f(x) or g(x)

Answer 7

so the top one is x^1/3 right

Answer 8

so itd be (x^1/3)^2 -(10-e^-.1x) dx bound 0 to 8?

Answer 9

jk the 10 one is first right

Answer 10

Yeah, the 10e^(-.1x) is the top function so your integral would be set up like this: \[\pi \int\limits _0 ^8 [(10e ^{-0.1x})^2-(x^{\frac{ 1 }{ 3 }})^2] dx\]

Answer 11

yay thank you so much!

Answer 12

No problem :)